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Math Help - rational continuity

  1. #1
    Senior Member sfspitfire23's Avatar
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    rational continuity

    Suppose that the function f : R → R is continuous and that f (x) = 0 if x is rational. Prove that f (x) = 0 for all x in R

    Now, I know that if f is continuous at x =a and {xn} is a sequence that converges to a, then { f (xn)} converges to f (a). But where does the rational part of the question come into play?
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    Chuse a sequence of rational numbers which converges to a.
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    If you like you could prove the more general: let f,g:X\to Y be continuous functions between topological spaces and suppose Y is Hausdorff. If f and g agree on a dense subset of X, they agree on all of X.
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    Suppose for some c that f(c) isn't equal to 0. We can assume, without loss of generality, that f(c) > 0. By continuity there is an open neighborhood about c such that f(x) > 0 for all x in that neighborhood. Now use the denseness of the rationals.
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    MHF Contributor Drexel28's Avatar
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    The below is more just a note of passing interest.




    Quote Originally Posted by maddas View Post
    If you like you could prove the more general: let f,g:X\to Y be continuous functions between topological spaces and suppose Y is Hausdorff. If f and g agree on a dense subset of X, they agree on all of X.
    Yes! The above is totally correct that is because if we define A(\varphi,\psi)=\left\{x:\varphi(x)=\psi(x)\right\  } and \varphi\oplus\psi:X\to Y\times Y by \varphi\oplus\psi:x\mapsto(\varphi(x),\psi(x)) then the continuity of \varphi,\psi easily implies the continuity of \varphi\oplus\psi. Thus, one notes that since Y is Hausdorff that \Delta\subseteq Y\times Y is closed and A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta) and so A(\varphi,\psi) is closed. So, if D\subseteq \subseteq A(\varphi,\psi)\subseteq X is dense then X=\overline{D}\subseteq\overline{A(\varphi,\psi)}=  A(\varphi,\psi) and so X=A(\varphi,\psi). But! By definition this means that \varphi(x)=\psi(x),\text{ }\forall x\in X or that \psi=\varphi!

    Maybe more intuitive is to notice that \mathbb{R} is a topological group (with the usual addition and usual topology). And since given a topological group G the map \alpha:G\times G\to G given by g,h)\mapsto gh^{-1}" alt="\alphag,h)\mapsto gh^{-1}" /> is continuous. Thus, one notes that \varphi\oplus\psi:G\to G\times G is continuous and thus so is \alpha\circ(\varphi\oplus\psi):G\to G with g\mapsto \varphi(g)(\psi(g))^{-1}. So, noticing that A(\varphi,\psi)=\left\{g\in G:\varphi(g)=\psi(g)\right\}=\left\{g\in G:\varphi(g)(\psi(g))^{-1}=e\right\} =\left(\alpha(\varphi\oplus\psi)\right)^{-1}(\{e\}) and thus the above claim is true if \{e\} is closed in G. And since \{0\} is closed in \mathbb{R} the conclusion follows!

    (notice that the above paragraph is a farce since \{e_G\}\text{ is closed }\implies G\text{ is }T_1\implies G\text{ is }T_2)
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    No need for a cannon when a flyswatter works.
    Theorem: If a continuous function f is either positive or negative at x=a then f(x) has the same sign on some open interval (a-\delta,a+\delta).
    So if f(b)\not= 0 then f(b) is either positive or negative .
    But in any interval about b there is a rational number.
    That contradicts the theorem. HOW?
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  7. #7
    Senior Member sfspitfire23's Avatar
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    but then there must be f(x)=0 b/c rationals are everywhere dense. So, f(x)=0 at least once
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  8. #8
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    Quote Originally Posted by Drexel28 View Post
    Yes! The above is totally correct that is because if we define A(\varphi,\psi)=\left\{x:\varphi(x)=\psi(x)\right\  } and \varphi\oplus\psi:X\to Y\times Y by \varphi\oplus\psi:x\mapsto(\varphi(x),\psi(x)) then the continuity of \varphi,\psi easily implies the continuity of \varphi\oplus\psi. Thus, one notes that since Y is Hausdorff that \Delta\subseteq Y\times Y is closed and A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta) and so A(\varphi,\psi) is closed. So, if D\subseteq \subseteq A(\varphi,\psi)\subseteq X is dense then X=\overline{D}\subseteq\overline{A(\varphi,\psi)}=  A(\varphi,\psi) and so X=A(\varphi,\psi). But! By definition this means that \varphi(x)=\psi(x),\text{ }\forall x\in X or that \psi=\varphi!

    Maybe more intuitive is to notice that \mathbb{R} is a topological group (with the usual addition and usual topology). And since given a topological group G the map \alpha:G\times G\to G given by g,h)\mapsto gh^{-1}" alt="\alphag,h)\mapsto gh^{-1}" /> is continuous. Thus, one notes that \varphi\oplus\psi:G\to G\times G is continuous and thus so is \alpha\circ(\varphi\oplus\psi):G\to G with g\mapsto \varphi(g)(\psi(g))^{-1}. So, noticing that A(\varphi,\psi)=\left\{g\in G:\varphi(g)=\psi(g)\right\}=\left\{g\in G:\varphi(g)(\psi(g))^{-1}=e\right\} =\left(\alpha(\varphi\oplus\psi)\right)^{-1}(\{e\}) and thus the above claim is true if \{e\} is closed in G. And since \{0\} is closed in \mathbb{R} the conclusion follows!

    (notice that the above paragraph is a farce since \{e_G\}\text{ is closed }\implies G\text{ is }T_1\implies G\text{ is }T_2)

    This is a fascinating proof. Its so algebraic :] I'm not sure I understand it all! I had thought of this low-tech one: suppose D is dense in X and there is a point a at which f and g differ. Use the Hausdorff property to pick disjoint open sets U and V containing f(a) and g(a) respectively. Then f^{-1}(U)\cap g^{-1}(V) is open in X so it meets D, at d say, and f(d)=g(d). But then f(d) is in both U and V, a contradiction.

    Here is something I have wondered before but been unable to prove: can we relax the condition that Y is T2? Are there distinct functions agreeing on a dense set that take values in a (say) T1 space? Do you know?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by maddas View Post

    Here is something I have wondered before but been unable to prove: can we relax the condition that Y is T2? Are there distinct functions agreeing on a dense set that take values in a (say) T1 space? Do you know?
    Hmm...

    My first inclination would be something like this.

    Let X be \mathbb{N} with the cofinite topology and define \varphi,\psi:X\to X by \varphi(x)=\begin{cases} x & \mbox{if}\quad x\ne 0 \\ 1 & \mbox{if}\quad x=0\end{cases}. Then, if O\subseteq X is open and thus infinite then \varphi^{-1}(O) is infinite and thus open. Similarly, define \psi(x) to be the identity map. Then, they agree on \mathbb{N}-\{0\} which is dense since given any point x\in X and any neighborhood of x that neighborhood must infinitely many points and thus a point different from both x and \{0\}.

    I hope I didn't make any stupid mistakes.
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